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What is the center of the ellipse 8x2+3y2−48=08x^2 + 3y^2 - 48 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 8x2+3y2−48=08x^2 + 3y^2 - 48 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 4848: Now, we need to get the equation in standard form by dividing everything by 4848.\newlinex2/(48/8)+y2/(48/3)=48/48x^2/(48/8) + y^2/(48/3) = 48/48\newlinex2/6+y2/16=1x^2/6 + y^2/16 = 1
  2. Standard Form Comparison: The standard form of an ellipse is (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k)(h, k) is the center.\newlineSo, comparing our equation to the standard form, we see that h=0h = 0 and k=0k = 0.\newlineTherefore, the center of the ellipse is (0,0)(0, 0).

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